Total Entropy and Entropy per trial 
I've attached here with a part from the Entropy chapter of Molecular Driving forces. The equations mentioned in the text are 
$$S=k\ln{w} \tag{5.1}$$
 and 
$$ S=-k\sum_{i=1}^{t}p_{i}\ln{p_{i}} \tag{5.2}$$
I haven't been able to absorb the result shown in (5.4). I can't understand how they got $\frac{S_{N}}{Nk}=-k\sum_{i=1}^{t}p_{i}\ln{p_{i}}$. The only way they could have got this one is by using $S_{N}=k\ln{w}$, but aren't $S_{N}$ and $S$ different. How could they use same relation $S_{N}=k\ln{w}$ and $S=k\ln{w}$? since both are different. One is total entropy while the another one is entropy per trial ?
 A: It's a bit unclear which part of it has had you confused, and I don't have access to your book, but from your equations $(5.1)$ and
$(5.2)$ I reckon the author is simply showing how the definition of
entropy in terms of log of multiplicity (number of microstates), can
be re-expressed in terms of a probability distribution. Where then you
usually take some example of coin tosses, or a box filled with coins,
gas in a box etc, and express the multiplicity $W$ of the problem, then
perform an Stirling approximation in order to get rid of the factorials
and deal only with powers of ratios of number times a result is
obtained over the total number of trials. 
Long story short, with the usual definition of entropy as $S = k_B
\ln{W}$ in mind, you get to the relation you've written:
$$
\ln{W} = - \sum_{i=1}^{t}n_i \ln{p_i} \tag{*}
$$ 
where I assume $n_i$ denotes the number of times some outcome $i$ has
occurred, $N$ total number of trials (of coin flip for instance) and
$n_i/N=p_i$ the probability of outcome $i$ (for instance heads). In
order to get to your $(5.2)$ equation, we still have to divide $(*)$ by
$N$ to go from $n_i \to p_i.$ 
$$
\frac{\ln{W}}{N} = -\sum_{i=1}^{t} p_i \ln{p_i} \overset{(5.2)}{=} \frac{S}{k_B}
$$
Now $S$ can be called the entropy per trial in this context, because it refers to the amount
of information one would gain after the revelation of the outcome of
one trial given there are $t$ types of different equi-probable possible outcomes. For example if we take the usual coin toss experiments ($t=2$), then with
the assumption that heads and tails are equi-probable, the outcome of
one toss carries only one bit of information or in terms of Shannon
entropy $\log_2{2}$ bits of information. Conversely, you
can do the interpretation the other way around, namely, in terms of the
amount of uncertainty attached with each toss (e.g. when trying to
predict an outcome). Finally, after $N$ tosses, if you ask what the
total entropy $S_N$ gain has been, then it's just $N$ times the amount
of entropy gained from each toss, meaning the total entropy is
extensive in the number of trials: $S_N = N*S.$ Which again you can interpret as the total amount of information you've gathered after repeating the experiment $N$ times.  
